- #1

kiuhnm

- 66

- 1

## Homework Statement

Suppose that a

*smooth*differential ##n-1##-form ##\omega## on ##\mathbb{R}^n## is ##0## outside of a ball of radius ##R##. Show that $$

\int_{\mathbb{R}^n} d\omega = 0.

$$

## Homework Equations

[/B]

$$\oint_{\partial K} \omega = \int_K d\omega$$

## The Attempt at a Solution

If ##\omega## is ##0## outside of the ball, by continuity, it must be ##0## on the surface of the ball as well. We know that $$

\omega = \sum_{i=1}^n f^i(x^1,\ldots,x^n) dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n

$$ for some functions ##f^i:\mathbb{R}^n\to\mathbb{R}##, so

$$

d\omega = \sum_{i=1}^n (-1)^{i-1} \frac{\partial f^i}{\partial x^i} dx^1\wedge\cdots\wedge dx^n.

$$

All those partial derivatives are ##0## outside of the ball, so $$

\int_{\mathbb{R}^n} d\omega = \int_B d\omega = \oint_{\partial B} \omega = 0,

$$ where ##B## is the ball.